i can't understand decorator
Oscar Benjamin
oscar.j.benjamin at gmail.com
Tue Jan 15 09:31:25 EST 2013
On 15 January 2013 14:20, contro opinion <contropinion at gmail.com> wrote:
>>>> def deco(func):
> ... def kdeco():
> ... print("before myfunc() called.")
> ... func()
> ... print(" after myfunc() called.")
> ... return kdeco
> ...
>>>> @deco
> ... def myfunc():
> ... print(" myfunc() called.")
> ...
>>>> myfunc()
> before myfunc() called.
> myfunc() called.
> after myfunc() called.
>>>> deco(myfunc)()
> before myfunc() called.
> before myfunc() called.
> myfunc() called.
> after myfunc() called.
> after myfunc() called.
> 1.
> why there are two lines :before myfunc() called.and tow lines :after
> myfunc() called. in the output?
You have wrapped the function twice with the decorator. Try changing the line
print("before func() called")
to
print("about to call", func,__name__)
and you'll see that the function it is about to call is not the same
in both cases.
> 2.why the result is not
> before myfunc() called.
> myfunc() called.
> after myfunc() called.
> before myfunc() called.
> myfunc() called.
> after myfunc() called.
You would get this output if you just called myfunc() twice. I don't
know why you expect wrapping the function twice to have this effect.
Oscar
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